Chapter 10 - Section 10.5 - Systems of Nonlinear Equations in Two Variables - Exercise Set - Page 810: 11

$(x,y)=${$(-5,-4),(3,0)$}

Re-write the second equation as: $x=2y+3$ First equation yields: $y^2=(2y+3)^2-9$ or, $3y(y+4)=0$ or, $y=${$-4,0$} From second equation when $y=-4$, we have $x=-5$ From second equation when $y=0$, we have $x=3$ Thus, solution set is $(x,y)=${$(-5,-4),(3,0)$}